Physics

CBSE Class 12

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11.

A ray of light, incident on an equilateral glass prism ( ) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.

Since the prism is equilateral in shape,

Angle of prism, A = 60^{o}

Angle of refraction, r = A/2 = 30^{o}

Now, using Snell’s law,

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12.

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self-inductance L is given by

Self-inductance of a coil is numerically equal to the magnetic flux linked with the coil when the current through coil is one Ampere.

Mathematically, it is given by,

where, L is the constant of proportionality and is called the self-inductance.

Energy stored in an inductor:

Consider a source of emf connected to an inductor L.

With increase in current, the opposing induced emf is given by,

If the source of emf sends a current i through the inductor for a small time dt, then the amount of work done by the source,

Hence, the total amount of work done by source of emf when the current increases from its initial values (i = 0) to its final value (I) is given by,

This work done gets stored in the inductor in the form of energy.

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13.

State de-Broglie hypothesis.

According to hypothesis of de Broglie "The atomic particles of matter moving with a given velocity, can display the wave like properties.

i.e.,

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14.

A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2A. A straight long wire carrying 5A current is kept near the loop as shown.

If the loop and the wire are coplanar, find

(i) the torque acting on the loop and

(ii) the magnitude and direction of the force on the loop due to the current carrying wire.

i) Torque on the current carrying conductor is given by,

In this case, M and B have the same direction. Therefore, = 0^{o}.

So Torque, = 0.

ii) The magnetic forces are equal and opposite on the parallel current carrying wires.

Therefore, force acting on the loop is given by,

The force is attractive in nature or is towards the conductor.

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15.

A metallic rod of ‘L’ length is rotated with angular frequency of ‘ ’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Emf is induced is given by,

where, is the rate of change of area of the loop formed by the sector OPQ.

At any instant of time t, let be the angle between the rod and the radius of the circle at P.

Area of the sector OPQ = ;R is the radius of the circle.

Therefore, emf induced is,

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16.

Distinguish between ‘Analog and Digital signals’.

Difference between analog and digital signals is:

Analog signal |
Digital signal |

They are continuous signals i.e., continuous variation of voltage or current. |
These are the signals which can take only discrete values. In digital signal, high means 1 and low means 0. |

E.g. Sound of a human |
E.g., Temperature |

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17.

Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 resistance. Also find the potential difference between A and D.

Applying Kirchhoff's loop rule for loop ABEFA,

–9 + 6 + 4 × 0 + 2I = 0

I = 1.5 A … (i)

For loop BCDEB,

3 + IR + 4 × 0 – 6 = 0

Therefore, IR = 3

Putting the value of I from equation (i), we have

Potential difference between A and D through path ABCD,

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18.

Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?

When immersed in a liquid, a bi-convex lens of glass will behave as a plane glass,

when, ; is the refractive index of liquid and is the refractive index of glass.
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19.

Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.

For a point charge Q,

Electric potential is inversely proportional to r and Electric field is inversely proportional to r^{2}.

Electric Potential |
Electric field |

The graph below shows us the variation of E and V with distance ‘r’.

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20.

The figure shows a series LCR circuit with L = 5.0 H, C = 80 mF, R = 40 W connected to a variable frequency 240V source. Calculate.

(i) The angular frequency of the source which drives the circuit at resonance.

(ii) The current at the resonating frequency.

(iii) The rms potential drop across the capacitor at resonance.

Given a series LCR circuit,

i) At resonance, angular frequency is given by,

ii) Current at resonating frequency is given by,

iii) RMS potential drop across the capacitor at resonance is given by,

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